Answer:
The one on the top left
Step-by-step explanation:
Slope of one third and y intercept -4
3(2+x)
open the parenthesis
= 6 + 3x
3(2+x) = 6 + 3x
Hence the statement of equality is true
H is the correct option
NOT MY WORDS TAKEN FROM A SOURCE!
(x^2) <64 => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2 so (x^2) - (8^2) < 0 To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y) You can work backwards and verify this is true. so let's set (x^2) - (8^2) equal to zero to find the roots: (x^2) - (8^2) = 0 => (x-8)*(x+8) = 0 if x-8 = 0 => x=8 and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0 so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0 so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0 so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.
Hope this helped!
:)
(8x^2-9y^2-4x)+(x^2-3y^2-7x)
First, get rid of the parentheses and combine like terms
(8x^2+x^2)(-9y^2-3y^2)(-4x-7x)
9x^2-12y^2-11x is your answer
To do this problem, you will need to know the side lengths of the triangle. If you have those, just plug them in for a, b, and c and evaluate the square root.
Here is an example. If the sides are 3, 4, 5.
It would be:
The 6 is in the formula because it is the semi-perimeter (half).
